The function f(x) = 2x^3 - 6x + 5 is an incre | vedclass

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(C) Given the function $f(x) = 2x^3 - 6x + 5$. To find the intervals where the function is increasing,we first find the derivative $f'(x)$. $f'(x) = \

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$x  1$

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(C) Given the function $f(x) = 2x^3 - 6x + 5$. To find the intervals where the function is increasing,we first find the derivative $f'(x)$. $f'(x) = \frac{d}{dx}(2x^3 - 6x + 5) = 6x^2 - 6$. $A$ function is increasing when $f'(x) > 0$. So,$6x^2 - 6 > 0$. Dividing by $6$,we get $x^2 - 1 > 0$,which factors as $(x - 1)(x + 1) > 0$. Using the sign scheme for the inequality $(x - 1)(x + 1) > 0$,the expression is positive when $x > 1$ or $x Thus,the function is increasing for $x \in (-\infty, -1) \cup (1, \infty)$.

The function $f(x) = 2x^3 - 6x + 5$ is an increasing function,if

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